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Problem Suppose that $(a_{j})_{j=0}^{\infty}$ is a sequence of positive numbers for which $a_{j+1}/a_{j} \to 1$, so that D'Alembert's test gives no information. Show that if

$$\limsup_{j\to\infty}\frac{j(a_{j+1}-a_{j})}{a_{j}}<0,$$ then, $\sum_{j=0}^{\infty}a_{j}$ converges.

This problem is Exercise 4.2.13 from A Course in Mathematical Analysis, Volume 1 (D. J. H. Garling). The previous exercise, Exercise 4.2.12, covered Kummer's convergence test, which is a test that looks like this.

Kummer's test Suppose that $(a_{j})_{j=0}^{\infty}$ and $(c_{j})_{j=0}^{\infty}$ are sequences of positive real numbers. If

$$\limsup_{j\to\infty}\left(\frac{c_{j+1}a_{j+1}}{a_{j}}-c_{j}\right)<0,$$

then $\sum_{j=0}^{\infty}a_{j}$ converges.

Exercise 4.2.13 says that it deals with a special case of kummer's test, I'm not sure how to connect it to kummer's test. Even the smallest hint would be very helpful.

Riemann
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gnahop
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    $\frac{j(a_{j+1}-a_{j})}{a_{j}}=\frac{ja_{j+1}}{a_{j}}-j$ and $(j)_{j=1}^\infty$ is a sequence of positive real numbers – M A Pelto Dec 21 '24 at 01:31
  • @MAPelto But, isn't it a contradiction to have $c_{j} = j, c_{j+1} = j$? – gnahop Dec 21 '24 at 01:46
  • Dang I misread $\frac{c_{j+1}a_{j+1}}{a_{j}}$ as $\frac{c_{j}a_{j+1}}{a_{j}}$, subscripts are small. How about use $j<j+1$. There done. – M A Pelto Dec 21 '24 at 02:01
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    This is false: for $a_j = 1/j,$ we get $ j \frac{a_{j+1}}{a_j} - j = \frac{j^2}{j+1} - j = \frac{-j}{j+1} \to -1.$ You need the limsup to be $< - 1$, in which case M A Pelto's argument applies (in Kummer's it's fine for the $c_j$ to only eventually all be positive (why?)). – stochasticboy321 Dec 21 '24 at 02:19
  • @stochasticboy321 good point. Since $\limsup_{j\to\infty} \frac{a_{j+1}}{a_j}=\lim_{j\to\infty} \frac{a_{j+1}}{a_j}=1$, the result should be about if either $\limsup_{j\to\infty}\frac{j(a_{j+1}-a_{j})}{a_{j}}<-1$ or (equivalently) $\limsup_{j\to\infty}\frac{(j+1)a_{j+1}-ja_{j}}{a_{j}}<0$. – M A Pelto Dec 21 '24 at 02:47
  • This is basically one part of Raabe's test. In this posting Raabe's and Bertrand's tests from Kummer's. – Mittens Jan 25 '25 at 21:14

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The correct statement should be

Theorem. Let $(a_j)_{j=0}^{\infty}$ be a sequence of positive numbers. If $\limsup_{j\to\infty}(\frac{j(a_{j+ 1}-a_{j})}{a_j})<-1$ and $\frac{a_{j+1}}{a_j}\to1$, then $\sum a_j$ converges.

Then the problem becomes a special case of Kummer's test.

Let $c_j=j$ and $(a_j)_{j=0}^{\infty}$ be a sequence of positive real numbers, for $j\in\mathbb{N}$, then we have $$\limsup_{j\to\infty}(\frac{c_{j+1}a_{j+ 1}}{a_j}-c_j)=\limsup_{j\to\infty}(\frac{(j+1)a_{j+1}}{a_j}-j)=\limsup_{j\to\infty}(\frac{j(a_{j+1}-a_j)}{a_j}+\frac{a_{j+1}}{a_j}).$$Then we mush show that $\limsup_{j\to\infty}(\frac{j(a_{j+1}-a_j)}{a_j}+\frac{a_{j+1}}{a_j})<0$.

We use the fact that for any bounded sequences $(u_n)_{n=0}^{\infty}$ and $(v_n)_{n=0}^{\infty}$,$$\limsup_{n\to\infty}(u_n+v_n) \leq \limsup_{n\to\infty}u_n+\limsup_{n\to\infty}v_n.$$ Hence $$\limsup_{j\to\infty}(\frac{j(a_{j+1}-a_j)}{a_j}+\frac{a_{j+1}}{a_j})\leq\limsup_{j\to\infty}\frac{j(a_{j+1}-a_j)}{a_j}+\limsup_{j\to\infty}\frac{a_{j+1}}{a_j}<-1+1=0.$$ Now we can apply the Kummer's test to conclude that $\sum a_j$ converges.

Jay
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